Assume that the trains arrive completely independently of one another (this implies, for instance, that one train arriving within a specific second doesn't exclude another train arriving at the same second, and therefore, theoretically, any number of trains can arrive within any given second). In that case, what we have is a so-called Poisson process.
To get there, let's start with your second-division. If the trains truly cannot arrive within the same second, but are otherwise independent, then we can just look at the $1200$ $1$-second intervals in a $20$-minute period, and ask whether a train arrived or not. This would then give us a binomial distribution, with $n = 1200$, and expected value of $2$. That means that $p$, the probability of "success" (i.e. "a train arriving") for every given second is $\frac{2}{1200} = \frac{1}{600}$. For instance, our calculations for (b) would be
$$
P(1\text{ train}) = \binom{1200}1\left(\frac{1}{600}\right)^1\left(\frac{599}{600}\right)^{1199}
$$
and in general, for $k$ trains we get
$$
P(k\text{ trains}) = \binom{1200}k\left(\frac{1}{600}\right)^k\left(\frac{599}{600}\right)^{1200 - k}
$$
Now, let's divide it even smaller. Say we divide it into milliseconds instead. Then we have new binomial distribution with $n = 1\,200\,000$ and still with expectation value $2$, which means that our new $p$ becomes $\frac{1}{600\,000}$. Now the probability of exactly one train arriving becomes
$$
P(1\text{ train}) = \binom{1\,200\,000}1\left(\frac{1}{600\,000}\right)^1 \left(\frac{599\,999}{600\,000}\right)^{1\,199\,999}
$$
and in general, for $k$ trains
$$
P(k\text{ trains}) = \binom{1\,200\,000}k\left(\frac{1}{600\,000}\right)^k \left(\frac{599\,999}{600\,000}\right)^{1\,200\,000-k}
$$
Continuing this way, we can, with a small leap of faith, see that the true distribution if the trains are completely independent from one another, is
$$
P(k \text{ trains}) = \lim_{n \to \infty}\binom{n}{k}\left(\frac{2}{n}\right)^k\left(1-\frac{2}{n}\right)^{n-k}
$$
This turns out to become $\dfrac{2^k\cdot e^{-2}}{k!}$, which is a simple enough formula that you can calculate it with most calculators. This is called the Poisson distribution (note that the $2$ that is raised to the $k$-th power and the $2$ in the exponential both come from the $2$ trains you expect within a $20$-minute period).
Best Answer
Your integrals are substantially overcounting the situations where they encounter each other, as well as including some situations where they shouldn't meet up at all. You can set it up in that fashion, but one integral should cover those situations where one friend arrives up to five minutes earlier than the other, and the other integral should cover the converse.
Actually, because you also have to take care of boundary situations, because neither friend ever shows up before 7:00. So, you'll end up with
$$ \int_{x=0}^5 \int_{y=0}^x \frac{1}{400} \, dy \, dx + \int_{x=5}^{20} \int_{y=x-5}^x \frac{1}{400} \, dy \, dx + \\ \int_{y=0}^5 \int_{x=0}^y \frac{1}{400} \, dy \, dx + \int_{y=5}^{20} \int_{x=y-5}^y \frac{1}{400} \, dy \, dx $$
It's a bit tedious, but should arrive at the proper answer.
The graphical method is to sketch out a square labelled 7:00 to 7:20 on the $x$-axis, and 7:00 to 7:20 on the $y$-axis. Each point in this square represents a time of arrival for one friend, and another time of arrival for the other friend. Shade in the area that represents situations where the friends meet. The fraction of the square that is shaded in is the probability you want.